pku2352 线段树

话说这题代码一小会就敲完了，但是说啥A不掉，一直wa，最后发现是边界为0我写了1，然后改了就对了。

题目大意：

   话说在一个平面直角坐标系有n个点，这n个点的等级的定义是它左下方向的点数，问从1~n-1每个等级各有多少个点。

简略题解：

   由于题目给的y第一x第二的升序，所以就用不着排序了，直接看x坐标，一个一个往里插，然后看比它小的有多少，用线段树维护。

（其实可以用平衡树，树状数组等等数据结构）

附程序（70行）：

program pku_2352; var ctree:array[0..70000] of record l,r,left,right,mid,sum:longint; end; level:array[0..15000] of longint; n,tot,sum:longint; {--------------------------------------} procedure build(x:longint); begin ctree[x].sum:=0; if ctree[x].l+1=ctree[x].r then exit; ctree[x].mid:=(ctree[x].l+ctree[x].r) div 2; inc(tot); ctree[tot].l:=ctree[x].l; ctree[tot].r:=ctree[x].mid; ctree[x].left:=tot; build(tot); inc(tot); ctree[tot].l:=ctree[x].mid; ctree[tot].r:=ctree[x].r; ctree[x].right:=tot; build(tot); end; {--------------------------------------} procedure insert(x,u:longint); begin inc(ctree[u].sum); if ctree[u].l+1=ctree[u].r then exit; if x>=ctree[u].mid then insert(x,ctree[u].right) else insert(x,ctree[u].left); end; {--------------------------------------} procedure find(x,u:longint); begin if ctree[u].l+1=ctree[u].r then begin if x>=ctree[u].l then inc(sum,ctree[u].sum); exit; end; if x>=ctree[u].r-1 then inc(sum,ctree[u].sum) else if x>=ctree[u].mid then begin find(x,ctree[u].left); find(x,ctree[u].right); end else find(x,ctree[u].left); end; {--------------------------------------} procedure main; var i,x,y:longint; begin fillchar(level,sizeof(level),0); ctree[1].l:=0;ctree[1].r:=33001; tot:=1; build(1); read(n); for i:=1 to n do begin read(x,y); sum:=0; find(x,1); inc(level[sum]); insert(x,1);end; for i:=0 to n-1 do writeln(level[i]); end; {--------------------------------------} begin main; end.